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Let $R$ be a Noetherian ring and $I$ an ideal of $R$. If $n$ is the cohomological dimension of $I$, then why is the following isomorphism true: $$H_{I}^{n}(M)\cong H_{I}^{n}(R)\otimes_R M.$$

The cohomological dimension of $I$ is defined to be the supremum of the set of integers $i$ such that $H_{I}^{i}(M)\neq 0$ for some $R$-module $M$.

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Also posted on MSE: math.stackexchange.com/questions/392162/… –  user23950 Jul 7 '13 at 8:16
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1 Answer 1

Assume $I$ is generated by $a_1,\dots,a_n$ (the cohomological dimension is $\ge$ the number of generators, so let's assume they are equal).

Let $C = C(a_1,\dots,a_n)$ be the Cech complex. Then, since $R$ is noetherian, for any complex $M$, we have that:

$R\Gamma_I M \cong C \otimes_R M$.

Thus, we want to calculate $H^n(C\otimes_R M)$.

As the cohomology of $C$ is supported in $[0,n]$, there is an isomorphism (See for example Lemma 15.3.6 of http://arxiv.org/pdf/1206.6632.pdf)

$H^n(C\otimes_R M) \cong H^n(C\otimes^L_R M) \cong H^n(C) \otimes_R H^0(M)$

And this is simply $H^n_I(R) \otimes_R M$.

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